If you don’t know what to do, think of what you could work out from the information in the question, and think of what information you would need to get to the answer. Since $ABCD$ is a square, we know that $\angle BDC$ is $45^\circ$. We also know that all the angles of an equilateral triangle are $60^\circ$.

Working backwards from what we need to find (the area of $\Delta DFC$), we know the base $DC=1$ so if we work out the altitude, we can find the area. Construct $G$ on $DC$ such that $FG\perp DC$. Let $x=FG$. \begin{align*} \tan\angle FDG &= \frac{FG}{DG}\\ \tan45^\circ &= \frac{x}{DG}\\ 1 &= \frac{x}{DG}\\ DG &= x \end{align*} (Or we could reason that $\angle DFG = \angle FDG$, so $\Delta DFG$ is isosceles and $DG = FG = x$.) \begin{align*} \tan\angle FCG &= \frac{FG}{GC}\\ \tan60^\circ &= \frac{x}{DC-DG}\\ \sqrt3 &= \frac{x}{1-x}\\ \sqrt3(1-x) &= x\\ \sqrt3 - \sqrt3 x &= x\\ \sqrt3 &= x+\sqrt3x\\ \sqrt3 &= (1+\sqrt3)x\\ x &= \frac{\sqrt3}{1+\sqrt 3} \end{align*} Then the area of $\Delta DFC$ is: \begin{align*} \frac{1}{2} DC \times FG &= \frac{1}{2} 1 \times x\\ &= \frac{\sqrt3}{2(1+\sqrt3)} \end{align*} which, if you prefer to rationalise the denominator, is $\frac{3 - \sqrt3}{4}$.